NCERT Solutions for Class 8 Maths - Cubes and Cube Roots - 1 (Exercise 6.1)

Q1: Which of the following numbers are not perfect cubes?
(i) 216 
(ii) 128 
(iii) 1000
(iv) 100 
(v) 46656

Sol: 

(i) We have 216 = 2 × 2 × 2 × 3 × 3 × 3Grouping the prime factors of 216 into triples, no factor is left over.∴ 216 is a perfect cube.

(ii) We have 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2Grouping the prime factors of 128 into triples, we are left with 2 as an ungrouped factor.∴ 128 is not a perfect cube.

(iii) We have 1000 = 2 × 2 × 2 × 5 × 5 × 5Grouping the prime factors of 1000 into triples, we are not left with any factor.∴ 1000 is a perfect cube.

(iv) We have 100 = 2 × 2 × 5 × 5Grouping the prime factors into triples, we do not get any triples. Factors 2 ×2 and 5 ×5 are not triples.∴ 100 is not a perfect cube.

(v) We have 46656 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3Grouping the prime factors of 46656 in triples, we are not leftover with any prime factor.∴ 46656 is a perfect cube.

Q2: Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube.
(i) 243 
(ii) 256 
(iii) 72
(iv) 675 
(v) 100
Sol: 

(i) We have 243 = 3 × 3 × 3 × 3 × 3The prime factor 3 is not a group of three.∴ 243 is not a perfect cube.
Now, [243] × 3 = [3 × 3 × 3 × 3 × 3] × 3
or 729 = 3 × 3 × 3 × 3 × 3 × 3
Now, 729 becomes a perfect cube.
Thus, the smallest required number to multiply 243 to make it a perfect cube is 3.

(ii) We have 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2Grouping the prime factors of 256 triples, we are left over with 2 × 2.∴ 256 is not a perfect cube.
Now, [256] × 2 = [2 × 2 × 2 × 2 × 2 × 2 × 2 × 2] × 2
or 512 = 2 ×2 ×2 ×2 ×2 ×2 ×2 ×2 × 2 i.e. 512 is a perfect cube.
Thus, the required smallest number is 2.

(iii) We have 72 = 2 × 2 × 2 × 3 × 3Grouping the prime factors of 72 in triples, we are left over with 3 ×3.∴ 72 is not a perfect cube.
Now, [72] × 3 = [2 × 2 × 2 × 3 × 3] × 3
or 216 = 2 × 2 × 2 × 3 × 3 × 3
i.e. 216 is a perfect cube.
∴ The smallest number required to multiply 72 to make it a perfect cube is 3.

(iv) We have 675 = 3 × 3 × 3 × 5 × 5Grouping the prime factors of 675 to triples, we are left over with 5 × 5
675 is not a perfect cube.
Now, [675] × 5 = [3 × 3 × 3 × 5 × 5] × 5
or 3375 = 3 × 3 × 3 × 5 × 5 × 5

Now, 3375 is a perfect cube.Thus, the smallest required number to multiply 675 such that the new number is a perfect cube is 5.

(v) We have 100 = 2 × 2 × 5 × 5The prime factor is not in the groups of triples.∴ 100 is not a perfect cube.
Now, [100] × 2 × 5 = [2 × 2 × 5 × 5] × 2 × 5
or [100] ×10 = 2 × 2 × 2 × 5 × 5 × 5
1000 = 2 × 2 × 2 × 5 × 5 × 5
Now, 1000 is a perfect cube.
Thus, the required smallest number is 10.

Q3: Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube.
(i) 81 
(ii) 128 
(iii) 135
(iv) 192 
(v) 704
Sol: 

(i) We have 81 = 3 × 3 × 3 × 3Grouping the prime factors of 81 into triples, we are left with 3.∴ 81 is not a perfect cube.
Now, [81] /3 = [3 × 3 × 3 × 3] ÷ 3
or  27 = 3 × 3 × 3
i.e. 27 is a perfect cube
Thus, the required smallest number is 3.

(ii) We have 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2Grouping the prime factors of 128 into triples, we are left with 2.∴ 128 is not a perfect cube
Now, [128] /2 = [2 × 2 × 2 × 2 × 2 × 2 × 2] ÷ 2
or  64 = 2 × 2 × 2 × 2 × 2 × 2
i.e. 64 is a perfect cube.
∴ The smallest required number is 2.

(iii) We have 135 = 3 × 3 × 3 × 5Grouping the prime factors of 135 into triples, we are left over with 5.∴ 135 is not a perfect cube
Now, [135] /5 = [3 × 3 × 3 × 5] ÷ 5
or 27 = 3 × 3 × 3
i.e. 27 is a perfect cube.
Thus, the required smallest number is 5.

(iv) We have 192 = 2 × 2 × 2 × 2 × 2 × 2 × 3Grouping the prime factors of 192 into triples, 3 is left over.∴ 192 is not a perfect cube.
Now, [192] / 3 = [2 × 2 × 2 × 2 × 2 × 2 × 3] ÷ 3
or  64 = 2 × 2 × 2 × 2 × 2 × 2
i.e. 64 is a perfect cube.
Thus, the required smallest number is 3.

(v) We have 704 = 2 × 2 × 2 × 2 × 2 × 2 × 11Grouping the prime factors of 704 into triples, 11 is left over.∴ 704 is not a perfect cube.
∴ [704] /11 = [2 × 2 × 2 × 2 × 2 × 2 × 11] ÷ 11
or 64 = 2 × 2 × 2 × 2 × 2 × 2
i.e. 64 is a perfect cube.
Thus, the required smallest number is 11.

Q4: Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube?

Sol: Sides of the cuboid are: 5 cm, 2 cm, 5 cm
∴ Volume of the cuboid = 5 cm × 2 cm × 5 cm
To form it as a cube, its dimension should be in the group of triples.

Given numbers = $5\; \backslash times\; 2\; \backslash times\; 5$5 × 2 × 5

Since, Factors of 5 and 2 both are not in group of three.
Therefore, the number must be multiplied by $2\; \backslash times\; 2\; \backslash times\; 5\; =\; 20$2 × 2 × 5 = 20 to make it a perfect cube.

Hence, he needs 20 cuboids.